Re: [Phys-L] electron velocity in an electric field.

[brian whatcott <betwys1@sbcglobal.net>]

On 4/24/2018 4:29 AM, David Bowman wrote:
> Regarding the answer to the question Brian W attempted to answer:
>
> > Q:Two parallel metal plates with a spacing of 1cm have a
> > potential difference of 10kv. An electron is projected from one
> > plate directly towards second.What is the initial velocity of
> > electron if it comes to rest just at the surface of second
> > plate?
> We assume the energy of the electron is conserved in its travels (i.e. we ignore the tiny radiative losses as the electron decelerates).  In this case the PE gain equals the initial KE lost in stopping the electron by the retarding electric field.  This means the initial KE of the electron is 10 keV.  To simplify the math let's keep things dimensionless.   So let  e = (KE)/(m*c^2), and let b = v/c.  This means
>
> e = 1/sqrt(1 - b^2) - 1 .
>
> Solving for b gives
>
> b = sqrt(2*e*(1 + e/2))/(1 + e).
>
> If we neglect the special relativistic corrections this formula simplifies to the Newtonian result
>
> b = sqrt(2*e).
>
> The special relativistic kinematic corrections themselves amount to a factor of
>
> sqrt(1 + e/2)/(1 + e).
>
> Now e = (10 keV)/(510.9989461(31) keV) = 0.01956951198(12) (CODATA 2014).
>
> This means
>
> b = 0.1949856095(6) with the relativisitic corrections, and
> b = 0.197835851 without them.
>
> Converting to SI units the initial speed is  v = b*c = 58,455.215 km/s.
>
> The relativistic correction is a factor of
> sqrt(1 + e/2)/(1 + e) = 0.985592897
> (i.e. about a 1.46% decrease from the Newtonian result).
>
> It stands to reason that the Newtonian result should be off by about 1 to 2 % since the initial kinetic energy is about 2% of the rest energy.
>
> BTW, using the New SI unit definitions the answer for the initial speed would be somewhat more precise than the 3.1*10^(-9) relative precision uncertainty given here because the electron mass/rest energy is more precisely known in terms of unified mass units than in terms of electron volts, and the new SI system (2018) numbers makes the conversions between unified mass units and kilograms or joules a matter of definition rather than measurement.
>
> David Bowman
> _______________________________________________
> Forum for Physics Educators
> Phys-l@mail.phys-l.org
> http://www.phys-l.org/mailman/listinfo/phys-l
David,

thanks to your kind contribution, I see my arithmetic in the first of 
three lines given below was in error

**************************

1/2 * 9.109 383 56 (11) × 10 −31 kg * v^2 = 10000 / 6.241*10^18

so v^2 = 10000/6.241*10^18 * 2 / ( 9.11* 10^-31) =0.730 *10^17

and V = 2.7 * 10^8 m/s !!!

**************************


1/2 * 9.109 X 10^-31 * v^2 = 10000 / 6.241 X 10^18

actually gives v^2 = 3.518*10^15 and v = 59313 km/s or

58483 km/s with the relativistic correction you gave

which agrees with your figure at the resolution I chose.

Other errors I noticed on rechecking: The breakdown voltage for air is 
around 3kV/mm (not 1 kV/mm as I offered); the breakdown voltage for 
(soft) vacuum only falls below 1 kV/mm at pressures between 0.3 and 10 
Torr judging by the Paschen curve for N2 given here:

https://en.wikipedia.org/wiki/Paschen%27s_law#/media/File:Paschen_curves.svg

(As remarked by Alex F Burr ~ priv. comm)

0.3 Torr is hardly beyond the capacity of a roughing vacuum pump, and 
just into the "fine" or "medium" pump range according to the table given 
here:

https://cas.web.cern.ch/sites/cas.web.cern.ch/files/lectures/platjadaro-2006/chew.pdf

...and  I would expect that a school lab could easily achieve this level.

And so this schoolhouse problem does appear to be capable of 
realization. (I noted in passing the similarity to Einstein's 
photoelectric experiment with clean copper sheets BTW)

With thanks

Brian W